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uitwerking oefeningen natuurkunde
- Instelling
- Universiteit Antwerpen (UA)
natuurkunde deel 1 - mechanica en thermodynamica
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CHAPTER 1: Introduction, Measurement, EstimatingResponses to Questions
1. (a) A particular person’s foot. Merits: reproducible. Drawbacks: not accessible to the general
public; not invariable (could change size with age, time of day, etc.); not indestructible.
(b) Any person’s foot. Merits: accessible. Drawbacks: not reproducible (different people have
different size feet); not invariable (could change size with age, time of day, etc.); not
indestructible.
Neither of these options would make a good standard.
2. The number of digits you present in your answer should represent the precision with which you
know a measurement; it says very little about the accuracy of the measurement. For example, if you
measure the length of a table to great precision, but with a measuring instrument that is not
calibrated correctly, you will not measure accurately.
3. The writers of the sign converted 3000 ft to meters without taking significant figures into account.
To be consistent, the elevation should be reported as 900 m.
4. The distance in miles is given to one significant figure and the distance in kilometers is given to five
significant figures! The figure in kilometers indicates more precision than really exists or than is
meaningful. The last digit represents a distance on the same order of magnitude as the car’s length!
5. If you are asked to measure a flower bed, and you report that it is “four,” you haven’t given enough
information for your answer to be useful. There is a large difference between a flower bed that is 4 m
long and one that is 4 ft long. Units are necessary to give meaning to the numerical answer.
6. Imagine the jar cut into slices each about the thickness of a marble. By looking through the bottom
of the jar, you can roughly count how many marbles are in one slice. Then estimate the height of the
jar in slices, or in marbles. By symmetry, we assume that all marbles are the same size and shape.
Therefore the total number of marbles in the jar will be the product of the number of marbles per
slice and the number of slices.
7. You should report a result of 8.32 cm. Your measurement had three significant figures. When you
multiply by 2, you are really multiplying by the integer 2, which is exact. The number of significant
figures is determined by your measurement.
8. The correct number of significant figures is three: sin 30.0º = 0.500.
9. You only need to measure the other ingredients to within 10% as well.
10. Useful assumptions include the population of the city, the fraction of people who own cars, the
average number of visits to a mechanic that each car makes in a year, the average number of weeks a
mechanic works in a year, and the average number of cars each mechanic can see in a week.
(a) There are about 800,000 people in San Francisco. Assume that half of them have cars. If each of
these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a
year. If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be
seen each week. Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a
week. The final estimate, then, is 800 car mechanics in San Francisco.
(b) Answers will vary.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
,Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
11. One common way is to observe Venus at a Sun
Venus
time when a line drawn from Earth to Venus
is perpendicular to a line connecting Venus
to the Sun. Then Earth, Venus, and the Sun
are at the vertices of a right triangle, with
Venus at the 90º angle. (This configuration
will result in the greatest angular distance
between Venus and the Sun, as seen from
Earth
Earth.) One can then measure the distance to
Venus, using radar, and measure the angular distance between Venus and the Sun. From this
information you can use trigonometry to calculate the length of the leg of the triangle that is the
distance from Earth to the Sun.
12. No. Length must be included as a base quantity.
Solutions to Problems
1. (a) 14 billion years = 1.4 × 1010 years
(b) (1.4 × 10 y )( 3.156 × 10 s 1 y ) =
10 7
4.4 × 1017 s
2. (a) 214 3 significant figures
(b) 81.60 4 significant figures
(c) 7.03 3 significant figures
(d) 0.03 1 significant figure
(e) 0.0086 2 significant figures
(f) 3236 4 significant figures
(g) 8700 2 significant figures
3. (a) 1.156 = 1.156 × 100
(b) 21.8 = 2.18 × 101
(c) 0.0068 = 6.8 × 10−3
(d) 328.65 = 3.2865 × 102
(e) 0.219 = 2.19 × 10−1
(f) 444 = 4.44 × 102
4. (a) 8.69 × 104 = 86, 900
(b) 9.1 × 103 = 9,100
(c) 8.8 × 10 −1 = 0.88
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
,Chapter 1 Introduction, Measurement, Estimating
(d) 4.76 × 10 2 = 476
(e) 3.62 × 10 −5 = 0.0000362
0.25 m
5. % uncertainty = × 100% = 4.6%
5.48 m
0.2 s
6. (a) % uncertainty = × 100% = 4%
5s
0.2 s
(b) % uncertainty = × 100% = 0.4%
50 s
0.2 s
(c) % uncertainty = × 100% = 0.07%
300 s
7. To add values with significant figures, adjust all values to be added so that their exponents are all the
same.
( ) ( ) ( ) ( ) (
9.2 × 103 s + 8.3 × 10 4 s + 0.008 × 106 s = 9.2 × 103 s + 83 × 103 s + 8 × 103 s ) ( )
= ( 9.2 + 83 + 8 ) × 103 s = 100.2 × 103 s = 1.00 × 105 s
When adding, keep the least accurate value, and so keep to the “ones” place in the last set of
parentheses.
8. ( 2.079 × 10 m )( 0.082 × 10 ) = 1.7 m . When multiplying, the result should have as many digits as
2 −1
the number with the least number of significant digits used in the calculation.
9. θ (radians) sin(θ ) tan(θ )
0 0.00 0.00 Keeping 2 significant figures in the angle, and
0.10 0.10 0.10 expressing the angle in radians, the largest angle that has
0.12 0.12 0.12 the same sine and tangent is 0.24 radians . In degrees,
0.20 0.20 0.20 the largest angle (keeping 2 significant figure) is 12°.
0.24 0.24 0.24 The spreadsheet used for this problem can be found on
0.25 0.25 0.26 the Media Manager, with filename
“PSE4_ISM_CH01.XLS,” on tab “Problem 1.9.”
10. To find the approximate uncertainty in the volume, calculate the volume for the minimum radius and
the volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the volume
is then half this variation in volume.
= 43 π ( 0.84 m ) = 2.483m 3
3
Vspecified = 43 π rspecified
3
= 43 π ( 0.80 m ) = 2.145 m 3
3
Vmin = 43 π rmin
3
= 43 π ( 0.88 m ) = 2.855 m 3
3
Vmax = 43 π rmax
3
ΔV = 1
2 (Vmax − Vmin ) = 12 ( 2.855 m3 − 2.145 m 3 ) = 0.355 m 3
ΔV 0.355 m 3
The percent uncertainty is = × 100 = 14.3 ≈ 14 % .
Vspecified 2.483 m 3
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3
, Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
11. (a) 286.6 mm 286.6 × 10−3 m 0.286 6 m
(b) 85 μ V 85 × 10−6 V 0.000 085 V
(c) 760 mg 760 × 10−6 kg 0.000 76 kg (if last zero is not significant)
(d) 60.0 ps 60.0 × 10 −12 s 0.000 000 000 060 0 s
(e) 22.5 fm 22.5 × 10 −15 m 0.000 000 000 000 022 5 m
(f) 2.50 gigavolts 2.5 × 109 volts 2, 500, 000, 000 volts
12. (a) 1 × 106 volts 1 megavolt = 1 Mvolt
(b) 2 × 10−6 meters 2 micrometers = 2μ m
(c) 6 × 103 days 6 kilodays = 6 kdays
(d) 18 ×102 bucks 18 hectobucks = 18 hbucks or 1.8 kilobucks
(e) 8 × 10−8 seconds 80 nanoseconds = 80 ns
13. Assuming a height of 5 feet 10 inches, then 5'10" = ( 70 in )(1 m 39.37 in ) = 1.8 m . Assuming a
weight of 165 lbs, then (165 lbs )( 0.456 kg 1 lb ) = 75.2 kg . Technically, pounds and mass
measure two separate properties. To make this conversion, we have to assume that we are at a
location where the acceleration due to gravity is 9.80 m/s2.
( )
14. (a) 93 million miles = 93 × 10 6 miles (1610 m 1 mile ) = 1.5 × 1011 m
(b) 1.5 × 10 m = 150 × 10 m = 150 gigameters or 1.5 × 1011 m = 0.15 × 1012 m = 0.15 terameters
11 9
0.111 yd 2
15. (a) 1 ft 2 = (1 ft 2 ) (1 yd 3 ft ) = 0.111 yd 2 , and so the conversion factor is
2
.
1 ft 2
10.8 ft 2
(b) 1 m 2 = 1 m 2( ) ( 3.28 ft 1 m )2 = 10.8 ft 2 , and so the conversion factor is 1m 2
.
16. Use the speed of the airplane to convert the travel distance into a time. d = vt , so t = d v .
t = d v = 1.00 km ⎜
⎛ 1 h ⎞ ⎛ 3600 s ⎞ = 3.8s
⎟⎜ ⎟
⎝ 950 km ⎠ ⎝ 1 h ⎠
(
17. (a) 1.0 × 10 −10 m = 1.0 × 10 −10 m ) ( 39.37 in 1 m ) = 3.9 × 10 −9 in
(1.0 cm ) ⎛⎜
1 m ⎞ ⎛ 1 atom ⎞
⎟ = 1.0 × 10 atoms
8
(b) ⎟⎜ −10
⎝ 100 cm ⎠ ⎝ 1.0 × 10 m ⎠
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4
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